Brother, they are talking about round robbin algorithm, where time quantum ='q' sec.
if you have n processes in ready queue with each have large burst time, the ready queue look like as : P1,P2,...Pn
now, if you run P1....how much time it will run? it should be 'q' sec, but it is not completed
therefore add P1 in ready queue, now ready queue look like as : P2,...Pn, P1
now analyze, next time P1 should run when after running of P2,P3,...Pn ====> Each process must wait no longer than (n − 1) × q time units its next time quantum.
what is the meaning of each process gets 1/n of the CPU time in chunks of at most q time units. ?
therefore in the time of n * q sec, each process can get only $\frac{1}{n}$ * q sec only.
for you example, 5 processes... P1,P2,P3,P4,P5 with tq=20 msec and assume every process requires more than 20 msec
Ready Queue:- P1,P2,P3,P4,P5
P1 |
|
0->20 ( after 20 sec P1 looses its turn ) |
|
Ready Queue:- P2,P3,P4,P5,P1
Ready Queue:- P3,P4,P5,P1,P2
P1 |
P2 |
P3 |
0->20 |
20->40 |
40->60 |
Ready Queue:- P4,P5,P1,P2,P3
P1 |
P2 |
P3 |
P4 |
0->20 |
20->40 |
40->60 |
60->80 |
Ready Queue:- P5,P1,P2,P3,P4
P1 |
P2 |
P3 |
P4 |
P5 |
0->20 |
20->40 |
40->60 |
60->80 |
80->100 |
Ready Queue:- P1,P2,P3,P4,P5 ==> ready queue is now look like as initial ==> P1 got its turn again after 80 msec => (n-1)*q
P1 |
P2 |
P3 |
P4 |
P5 |
0->20 |
20->40 |
40->60 |
60->80 |
80->100 |
total of 100 msec, each process get only 20 msec ==> therefore in the time of n * q sec, each process can get only $\frac{1}{n}$ * q sec only.